# Complicate density problem

## Homework Statement

We are lowering two small balls into the container filled with oil. We are measuring the final velocities of these balls achieved. For the steel ball we measure the final velocity of v1 = 0.23m/s and for the aluminum ball we measure v2 = 0.06m/ s. What are the density and the viscosity of oil, if the radius's of both balls are r = 3mm; the density of steel is ρ1 = 7800 kg/m3, and the density of aluminum is ρ2 = 2700 kg/m3. Assume that the linear law of resistance applies.

ρ= m/V

## The Attempt at a Solution

Volume of the sphere= 4/3 πr³
1000 kg/m³= 1 g/cm³
r= 3 mm= 0.3 cm
ρ= 7800 kg/m³= 7.8 g/cm³
ρ= 2700 kg/m³= 2.7 g/cm³

V= 0.11304 cm³

ρ = m/V → m= ρV

m(steal ball)= 0.88 g
m(aluminum ball)= 0.31 g

And this is all I managed to calculate. I know that conservation of mass and conservation of momentum apply to the density problems, but I don’t know how to apply them to find the density of oil.
Thank you!

Borek
Mentor
Calculation for the steal ball:

Fd= Fg
6πμrv(1)= m(1)g
μ= m(1)g/ 6πrv(1)
μ= 666.81 Pa s

Calculation for the aluminum ball:

Fd= Fg
6πμrv(2)= m(2)g
μ= m(2)g/ 6πrv(2)
μ= 1045.54 Pa s

So I calculated the viscosity according to Stokes' equation, however I had two balls in the question, one of steel and the other of aluminum so I just went ahead and plugged in both of them in seperate equations, and that gave me two different vicosities.

How do I continue?

Borek
Mentor
You can't calculate viscosity not knowing density.

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Can you please, give me another hint for how to approach my problem?
I'm lost and I would really like to understand the problem and find the solution.

You use the Stoke law n(viscosity)=[2(density of steel-density of oil)*g*R]/(9v1)
v1 is the velocity for steel
Again use the same formula for The case of aluminium and divide one equation with other..so that only the unknown will be the density of oil

Borek
Mentor
In other words - you have two equations in two unknowns.

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methods

OK, I got the final formula:

ρ (oil)= [ρ(steal)v(aluminum) – ρ(aluminum)v(steal)] / [v(aluminum) – v(steal)]
ρ(oil)= 900 kg/m³

And now for viscosity:

μ= 2/9 [ρ(steal) – ρ(oil) / v(steal)] gr²
μ= 5.88 Pa s

Hopefully, my calculations are now correct!?:shy:
Thank you for helping!